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2

We have \begin{align} \frac{\pi}{4} &= \int_{0}^{1}\frac{dt}{1 + t^{2}}\notag\\ &= \int_{0}^{1}\left(1 - t^{2} + \cdots + (-1)^{n - 1}t^{2n - 2} + (-1)^{n}\frac{t^{2n}}{1 + t^{2}}\right)\,dt\notag\\ &= 1 - \frac{1}{3} + \cdots + (-1)^{n - 1}\frac{1}{2n - 1} + (-1)^{n}\int_{0}^{1}\frac{t^{2n}}{1 + t^{2}}\,dt\notag\\ &= S_{n} + ...

10

A probability distribution of the continued fraction expansion terms follows the Gauss-Kuzmin distribution for almost all irrational numbers: $$p(k)=-\log_2\left(1-\frac{1}{(k+1)^2}\right)=\log_2\frac{(k+1)^2}{k(k+2)}$$ All generalized Khinchin's constants (including $K=K_0$, the geometric mean), are derived from this distribution. In this case, you seek ...

5

By Stirling's approximation, we know that $(n!)^{1/n} \sim n/e$ as $n\to\infty$, so $$\left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{1/n} \;=\; \left( \left(\left\lfloor \frac{n}{3} \right\rfloor!\right)^{3/n}\right)^{1/3} \;\sim\; \left(\frac{n}{3e}\right)^{1/3},$$ and hence $$f(n) \;\sim\; \left(\frac{2n}{3e}\right)^{1/3}$$ as $n\to\infty$.

3

Stirling's formula states that $n! \sim \sqrt{2\pi n} (n/e)^n$, and in particular $n!^{1/n} \sim n/e$. Therefore $$f(n) = (n/3)!^{1/n} (2^{n/3+1})^{1/n} \sim (n/3e)^{1/3} 2^{1/3} = \sqrt[3]{\frac{2n}{3e}}.$$

0

Your continued fraction is very generic, in fact it specifies only $c_0 = 0$ and $b_1 = 1$ (compare with Batominovski's expansion) the general form $$x = \frac{1}{a} = c_0 + \cfrac{b_1}{c_1+\cfrac{b_2}{c_2+\cfrac{b_3}{c_3+\cdots}}}$$ How to visualize such a limit process? For specific values of the coefficients $a_i$ and $b_i$ one might plot the ...

0

What about $\cfrac{1}{a}=\cfrac{1}{0+\cfrac{l}{m+\cfrac{n}{o+\cfrac{p}{q+\ldots}}}}$? In fact, I don't know what you mean by "visualization."

1

(A partial answer.) I tested your cfrac with the order 12 discussed by Naika (which in turn is a special case of a general cfrac by Ramanujan) and labeled as $D_1(q)$ here, D_1(q)= \dfrac{q(1-q)} {1-q^3+\dfrac{q^3(1-q^2)(1-q^4)} {(1-q^3)(1+q^6)+\dfrac{q^3(1-q^8)(1-q^{10})} {(1-q^3)(1+q^{12})+\dfrac{q^3(1-q^{14})(1-q^{16})} {(1-q^3)(1+q^{18})+\ddots }}}} ...

1

It works like this: $\frac{1}{a_1x+\frac{1}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}=\frac{\frac{1}{a_1x}}{1+\frac{\frac{1}{a_1x}}{b_1+\frac{1}{a_2x+\frac{1}{b_2}}}}$, because we multiply the top and bottom of the outermost fraction by $\frac{1}{a_1x}$. Now, isolate the fraction from the denominator above: ...

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